How many kwh of energy does a 650 −w toaster use in the morning if it is in operation for a total of 15 min ?

Two cats are sitting on a 30 m hill. One cat

V125BC [204]

Answer:

Cat with 6 kg sitting on a 30 meter hill has greater potential energy.

Explanation:

gravitational potential energy (J) = mass (kg) × gravitational field × height (m)

gravitational field in Earth is 10 N/kg = <u>fixed unit</u>

<h3 /><h3><u>Given</u>:</h3>

1st Cat: 4 kg

2nd Cat: 6 kg

Both of their height is 30 meters.

<h3><u>Solve for g.p.e</u>:</h3>

1st cat = mgh = 4 × 10 × 30 = 1200 Joules

2nd cat = mgh = 6 × 10 × 30 = 1800 Joules

3 0

2 years ago

The intensity of a sound wave increases as the wave spreads out from the source of the sound. True or false

ludmilkaskok [199]

Answer:False

Explanation:

the intensity of the sound wave decreases with increasing distance from the source.

3 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$